Convex Optimization — Boyd & Vandenberghe
5. Duality
• Lagrange dual problem
• weak and strong duality
• geometric interpretation
• optimality conditions
• perturbation and sensitivity analysis
• examples
• generalized inequalities
5–1
Lagrangian
standard form problem (not necessarily convex)
minimize f0(x)
subject to fi(x) ≤ 0,
hi(x) = 0,
i = 1, . . . , m
i = 1, . . . , p
variable x ∈ Rn, domain D, optimal value p⋆
Lagrangian: L : Rn × Rm × Rp → R, with dom L = D × Rm × Rp,
L(x, λ, ν) = f0(x) +
m
X
i=1
λifi(x) +
p
X
νihi(x)
i=1
• weighted sum of objective and constraint functions
• λi is Lagrange multiplier associated with fi(x) ≤ 0
• νi is Lagrange multiplier associated with hi(x) = 0
Duality
5–2
Lagrange dual function
Lagrange dual function: g : Rm × Rp → R,
g(λ, ν) =
=
inf L(x, λ, ν)
x∈D
inf
x∈D
f0(x) +
m
X
λifi(x) +
i=1
p
X
νihi(x)
i=1
!
g is concave, can be −∞ for some λ, ν
lower bound property: if λ 0, then g(λ, ν) ≤ p⋆
proof: if x̃ is feasible and λ 0, then
f0(x̃) ≥ L(x̃, λ, ν) ≥ inf L(x, λ, ν) = g(λ, ν)
x∈D
minimizing over all feasible x̃ gives p⋆ ≥ g(λ, ν)
Duality
5–3
Least-norm solution of linear equations
minimize xT x
subject to Ax = b
dual function
• Lagrangian is L(x, ν) = xT x + ν T (Ax − b)
• to minimize L over x, set gradient equal to zero:
∇xL(x, ν) = 2x + AT ν = 0
=⇒
x = −(1/2)AT ν
• plug in in L to obtain g:
1
g(ν) = L((−1/2)AT ν, ν) = − ν T AAT ν − bT ν
4
a concave function of ν
lower bound property: p⋆ ≥ −(1/4)ν T AAT ν − bT ν for all ν
Duality
5–4
Standard form LP
minimize cT x
subject to Ax = b,
x0
dual function
• Lagrangian is
L(x, λ, ν) = cT x + ν T (Ax − b) − λT x
= −bT ν + (c + AT ν − λ)T x
• L is affine in x, hence
g(λ, ν) = inf L(x, λ, ν) =
x
−bT ν
−∞
AT ν − λ + c = 0
otherwise
g is linear on affine domain {(λ, ν) | AT ν − λ + c = 0}, hence concave
lower bound property: p⋆ ≥ −bT ν if AT ν + c 0
Duality
5–5
Equality constrained norm minimization
minimize kxk
subject to Ax = b
dual function
T
T
g(ν) = inf (kxk − ν Ax + b ν) =
x
bT ν kAT νk∗ ≤ 1
−∞ otherwise
where kvk∗ = supkuk≤1 uT v is dual norm of k · k
proof: follows from inf x(kxk − y T x) = 0 if kyk∗ ≤ 1, −∞ otherwise
• if kyk∗ ≤ 1, then kxk − y T x ≥ 0 for all x, with equality if x = 0
• if kyk∗ > 1, choose x = tu where kuk ≤ 1, uT y = kyk∗ > 1:
kxk − y T x = t(kuk − kyk∗) → −∞
as t → ∞
lower bound property: p⋆ ≥ bT ν if kAT νk∗ ≤ 1
Duality
5–6
Two-way partitioning
minimize xT W x
subject to x2i = 1,
i = 1, . . . , n
• a nonconvex problem; feasible set contains 2n discrete points
• interpretation: partition {1, . . . , n} in two sets; Wij is cost of assigning
i, j to the same set; −Wij is cost of assigning to different sets
dual function
T
g(ν) = inf (x W x +
x
X
i
νi(x2i − 1)) = inf xT (W + diag(ν))x − 1T ν
x
−1T ν W + diag(ν) 0
=
−∞
otherwise
lower bound property: p⋆ ≥ −1T ν if W + diag(ν) 0
example: ν = −λmin(W )1 gives bound p⋆ ≥ nλmin(W )
Duality
5–7
Lagrange dual and conjugate function
minimize f0(x)
subject to Ax b,
Cx = d
dual function
g(λ, ν) =
inf
x∈dom f0
T
T
T
T
T
f0(x) + (A λ + C ν) x − b λ − d ν
= −f0∗(−AT λ − C T ν) − bT λ − dT ν
• recall definition of conjugate f ∗(y) = supx∈dom f (y T x − f (x))
• simplifies derivation of dual if conjugate of f0 is known
example: entropy maximization
f0(x) =
n
X
i=1
Duality
xi log xi,
f0∗(y) =
n
X
eyi−1
i=1
5–8
The dual problem
Lagrange dual problem
maximize g(λ, ν)
subject to λ 0
• finds best lower bound on p⋆, obtained from Lagrange dual function
• a convex optimization problem; optimal value denoted d⋆
• λ, ν are dual feasible if λ 0, (λ, ν) ∈ dom g
• often simplified by making implicit constraint (λ, ν) ∈ dom g explicit
example: standard form LP and its dual (page 5–5)
minimize cT x
subject to Ax = b
x0
Duality
maximize −bT ν
subject to AT ν + c 0
5–9
Weak and strong duality
weak duality: d⋆ ≤ p⋆
• always holds (for convex and nonconvex problems)
• can be used to find nontrivial lower bounds for difficult problems
for example, solving the SDP
maximize −1T ν
subject to W + diag(ν) 0
gives a lower bound for the two-way partitioning problem on page 5–7
strong duality: d⋆ = p⋆
• does not hold in general
• (usually) holds for convex problems
• conditions that guarantee strong duality in convex problems are called
constraint qualifications
Duality
5–10
Slater’s constraint qualification
strong duality holds for a convex problem
minimize f0(x)
subject to fi(x) ≤ 0,
Ax = b
i = 1, . . . , m
if it is strictly feasible, i.e.,
∃x ∈ int D :
fi(x) < 0,
i = 1, . . . , m,
Ax = b
• also guarantees that the dual optimum is attained (if p⋆ > −∞)
• can be sharpened: e.g., can replace int D with relint D (interior
relative to affine hull); linear inequalities do not need to hold with strict
inequality, . . .
• there exist many other types of constraint qualifications
Duality
5–11
Inequality form LP
primal problem
minimize cT x
subject to Ax b
dual function
T
T
T
g(λ) = inf (c + A λ) x − b λ =
x
dual problem
−bT λ AT λ + c = 0
−∞
otherwise
maximize −bT λ
subject to AT λ + c = 0,
λ0
• from Slater’s condition: p⋆ = d⋆ if Ax̃ ≺ b for some x̃
• in fact, p⋆ = d⋆ except when primal and dual are infeasible
Duality
5–12
Quadratic program
primal problem (assume P ∈ Sn++)
minimize xT P x
subject to Ax b
dual function
1
g(λ) = inf x P x + λ (Ax − b) = − λT AP −1AT λ − bT λ
x
4
T
T
dual problem
maximize −(1/4)λT AP −1AT λ − bT λ
subject to λ 0
• from Slater’s condition: p⋆ = d⋆ if Ax̃ ≺ b for some x̃
• in fact, p⋆ = d⋆ always
Duality
5–13
A nonconvex problem with strong duality
minimize xT Ax + 2bT x
subject to xT x ≤ 1
A 6 0, hence nonconvex
dual function: g(λ) = inf x(xT (A + λI)x + 2bT x − λ)
• unbounded below if A + λI 6 0 or if A + λI 0 and b 6∈ R(A + λI)
• minimized by x = −(A + λI)†b otherwise: g(λ) = −bT (A + λI)†b − λ
dual problem and equivalent SDP:
maximize −bT (A + λI)†b − λ
subject to A + λI 0
b ∈ R(A + λI)
maximize
−t
−λ
A + λI
subject to
bT
b
t
0
strong duality although primal problem is not convex (not easy to show)
Duality
5–14
Geometric interpretation
for simplicity, consider problem with one constraint f1(x) ≤ 0
interpretation of dual function:
g(λ) =
inf (t + λu),
(u,t)∈G
where
G = {(f1(x), f0(x)) | x ∈ D}
t
t
G
G
p⋆
d⋆
p⋆
λu + t = g(λ)
g(λ)
u
u
• λu + t = g(λ) is (non-vertical) supporting hyperplane to G
• hyperplane intersects t-axis at t = g(λ)
Duality
5–15
epigraph variation: same interpretation if G is replaced with
A = {(u, t) | f1(x) ≤ u, f0(x) ≤ t for some x ∈ D}
t
A
λu + t = g(λ)
p⋆
g(λ)
u
strong duality
• holds if there is a non-vertical supporting hyperplane to A at (0, p⋆)
• for convex problem, A is convex, hence has supp. hyperplane at (0, p⋆)
• Slater’s condition: if there exist (ũ, t̃) ∈ A with ũ < 0, then supporting
hyperplanes at (0, p⋆) must be non-vertical
Duality
5–16
Complementary slackness
assume strong duality holds, x⋆ is primal optimal, (λ⋆, ν ⋆) is dual optimal
f0(x⋆) = g(λ⋆, ν ⋆) = inf
x
f0(x) +
≤ f0(x⋆) +
m
X
i=1
m
X
λ⋆ifi(x⋆) +
i=1
≤ f0(x⋆)
λ⋆ifi(x) +
p
X
νi⋆hi(x)
i=1
p
X
!
νi⋆hi(x⋆)
i=1
hence, the two inequalities hold with equality
• x⋆ minimizes L(x, λ⋆, ν ⋆)
• λ⋆ifi(x⋆) = 0 for i = 1, . . . , m (known as complementary slackness):
λ⋆i > 0 =⇒ fi(x⋆) = 0,
Duality
fi(x⋆) < 0 =⇒ λ⋆i = 0
5–17
Karush-Kuhn-Tucker (KKT) conditions
the following four conditions are called KKT conditions (for a problem with
differentiable fi, hi):
1. primal constraints: fi(x) ≤ 0, i = 1, . . . , m, hi(x) = 0, i = 1, . . . , p
2. dual constraints: λ 0
3. complementary slackness: λifi(x) = 0, i = 1, . . . , m
4. gradient of Lagrangian with respect to x vanishes:
∇f0(x) +
m
X
i=1
λi∇fi(x) +
p
X
νi∇hi(x) = 0
i=1
from page 5–17: if strong duality holds and x, λ, ν are optimal, then they
must satisfy the KKT conditions
Duality
5–18
KKT conditions for convex problem
if x̃, λ̃, ν̃ satisfy KKT for a convex problem, then they are optimal:
• from complementary slackness: f0(x̃) = L(x̃, λ̃, ν̃)
• from 4th condition (and convexity): g(λ̃, ν̃) = L(x̃, λ̃, ν̃)
hence, f0(x̃) = g(λ̃, ν̃)
if Slater’s condition is satisfied:
x is optimal if and only if there exist λ, ν that satisfy KKT conditions
• recall that Slater implies strong duality, and dual optimum is attained
• generalizes optimality condition ∇f0(x) = 0 for unconstrained problem
Duality
5–19
example: water-filling (assume αi > 0)
Pn
minimize − i=1 log(xi + αi)
subject to x 0, 1T x = 1
x is optimal iff x 0, 1T x = 1, and there exist λ ∈ Rn, ν ∈ R such that
λ 0,
λixi = 0,
1
+ λi = ν
xi + α i
• if ν < 1/αi: λi = 0 and xi = 1/ν − αi
• if ν ≥ 1/αi: λi = ν − 1/αi and xi = 0
Pn
• determine ν from 1T x = i=1 max{0, 1/ν − αi} = 1
interpretation
• n patches; level of patch i is at height αi
1/ν ⋆
xi
• flood area with unit amount of water
• resulting level is 1/ν
Duality
αi
⋆
i
5–20
Perturbation and sensitivity analysis
(unperturbed) optimization problem and its dual
minimize f0(x)
subject to fi(x) ≤ 0,
hi(x) = 0,
maximize g(λ, ν)
subject to λ 0
i = 1, . . . , m
i = 1, . . . , p
perturbed problem and its dual
min. f0(x)
s.t. fi(x) ≤ ui,
hi(x) = vi,
i = 1, . . . , m
i = 1, . . . , p
max. g(λ, ν) − uT λ − v T ν
s.t.
λ0
• x is primal variable; u, v are parameters
• p⋆(u, v) is optimal value as a function of u, v
• we are interested in information about p⋆(u, v) that we can obtain from
the solution of the unperturbed problem and its dual
Duality
5–21
global sensitivity result
assume strong duality holds for unperturbed problem, and that λ⋆, ν ⋆ are
dual optimal for unperturbed problem
apply weak duality to perturbed problem:
p⋆(u, v) ≥ g(λ⋆, ν ⋆) − uT λ⋆ − v T ν ⋆
= p⋆(0, 0) − uT λ⋆ − v T ν ⋆
sensitivity interpretation
• if λ⋆i large: p⋆ increases greatly if we tighten constraint i (ui < 0)
• if λ⋆i small: p⋆ does not decrease much if we loosen constraint i (ui > 0)
• if νi⋆ large and positive: p⋆ increases greatly if we take vi < 0;
if νi⋆ large and negative: p⋆ increases greatly if we take vi > 0
• if νi⋆ small and positive: p⋆ does not decrease much if we take vi > 0;
if νi⋆ small and negative: p⋆ does not decrease much if we take vi < 0
Duality
5–22
local sensitivity: if (in addition) p⋆(u, v) is differentiable at (0, 0), then
λ⋆i
∂p⋆(0, 0)
=−
,
∂ui
νi⋆
∂p⋆(0, 0)
=−
∂vi
proof (for λ⋆i): from global sensitivity result,
∂p⋆(0, 0)
p⋆(tei, 0) − p⋆(0, 0)
= lim
≥ −λ⋆i
tց0
∂ui
t
p⋆(tei, 0) − p⋆(0, 0)
∂p⋆(0, 0)
= lim
≤ −λ⋆i
tր0
∂ui
t
hence, equality
p⋆(u) for a problem with one (inequality)
constraint:
u
p⋆(u)
u=0
p⋆(0) − λ⋆u
Duality
5–23
Duality and problem reformulations
• equivalent formulations of a problem can lead to very different duals
• reformulating the primal problem can be useful when the dual is difficult
to derive, or uninteresting
common reformulations
• introduce new variables and equality constraints
• make explicit constraints implicit or vice-versa
• transform objective or constraint functions
e.g., replace f0(x) by φ(f0(x)) with φ convex, increasing
Duality
5–24
Introducing new variables and equality constraints
minimize f0(Ax + b)
• dual function is constant: g = inf x L(x) = inf x f0(Ax + b) = p⋆
• we have strong duality, but dual is quite useless
reformulated problem and its dual
minimize f0(y)
subject to Ax + b − y = 0
maximize bT ν − f0∗(ν)
subject to AT ν = 0
dual function follows from
g(ν) = inf (f0(y) − ν T y + ν T Ax + bT ν)
x,y
−f0∗(ν) + bT ν AT ν = 0
=
−∞
otherwise
Duality
5–25
norm approximation problem: minimize kAx − bk
minimize kyk
subject to y = Ax − b
can look up conjugate of k · k, or derive dual directly
g(ν) = inf (kyk + ν T y − ν T Ax + bT ν)
x,y
T
b ν + inf y (kyk + ν T y) AT ν = 0
=
−∞
otherwise
T
b ν AT ν = 0, kνk∗ ≤ 1
=
−∞ otherwise
(see page 5–4)
dual of norm approximation problem
maximize bT ν
subject to AT ν = 0,
Duality
kνk∗ ≤ 1
5–26
Implicit constraints
LP with box constraints: primal and dual problem
minimize cT x
subject to Ax = b
−1 x 1
maximize −bT ν − 1T λ1 − 1T λ2
subject to c + AT ν + λ1 − λ2 = 0
λ1 0, λ2 0
reformulation with box constraints made implicit
T
c x −1 x 1
minimize f0(x) =
∞
otherwise
subject to Ax = b
dual function
g(ν) =
inf
(cT x + ν T (Ax − b))
−1x1
= −bT ν − kAT ν + ck1
dual problem: maximize −bT ν − kAT ν + ck1
Duality
5–27
Problems with generalized inequalities
minimize f0(x)
subject to fi(x) Ki 0, i = 1, . . . , m
hi(x) = 0, i = 1, . . . , p
Ki is generalized inequality on Rki
definitions are parallel to scalar case:
• Lagrange multiplier for fi(x) Ki 0 is vector λi ∈ Rki
• Lagrangian L : Rn × Rk1 × · · · × Rkm × Rp → R, is defined as
L(x, λ1, · · · , λm, ν) = f0(x) +
m
X
i=1
λTi fi(x) +
p
X
νihi(x)
i=1
• dual function g : Rk1 × · · · × Rkm × Rp → R, is defined as
g(λ1, . . . , λm, ν) = inf L(x, λ1, · · · , λm, ν)
x∈D
Duality
5–28
lower bound property: if λi Ki∗ 0, then g(λ1, . . . , λm, ν) ≤ p⋆
proof: if x̃ is feasible and λ Ki∗ 0, then
f0(x̃) ≥ f0(x̃) +
m
X
λTi fi(x̃) +
i=1
≥
p
X
νihi(x̃)
i=1
inf L(x, λ1, . . . , λm, ν)
x∈D
= g(λ1, . . . , λm, ν)
minimizing over all feasible x̃ gives p⋆ ≥ g(λ1, . . . , λm, ν)
dual problem
maximize g(λ1, . . . , λm, ν)
subject to λi Ki∗ 0, i = 1, . . . , m
• weak duality: p⋆ ≥ d⋆ always
• strong duality: p⋆ = d⋆ for convex problem with constraint qualification
(for example, Slater’s: primal problem is strictly feasible)
Duality
5–29
Semidefinite program
primal SDP (Fi, G ∈ Sk )
minimize cT x
subject to x1F1 + · · · + xnFn G
• Lagrange multiplier is matrix Z ∈ Sk
• Lagrangian L(x, Z) = cT x + tr (Z(x1F1 + · · · + xnFn − G))
• dual function
g(Z) = inf L(x, Z) =
x
− tr(GZ) tr(FiZ) + ci = 0,
−∞
otherwise
i = 1, . . . , n
dual SDP
maximize − tr(GZ)
subject to Z 0, tr(FiZ) + ci = 0,
i = 1, . . . , n
p⋆ = d⋆ if primal SDP is strictly feasible (∃x with x1F1 + · · · + xnFn ≺ G)
Duality
5–30